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Difference between revisions of "2010 AMC 10A Problems/Problem 23"

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Using the first few values of <math>n</math>, it is easy to derive a formula for <math>P(n)</math>. The chance that she stops on the second box (<math>n=2</math>) is the chance of drawing a white marble then a red marble: <math>\frac{1}2 * \frac{1}3</math>. The chance that she stops on the third box (<math>n=3</math>) is the chance of drawing two white marbles then a red marble:<math>\frac{1}2 * \frac{2}3 * \frac{1}4</math>. If <math>n=4</math>, <math>P(n) = \frac{1}2 * \frac{2}3 * \frac{3}4 * \frac{1}5</math>.
 
Using the first few values of <math>n</math>, it is easy to derive a formula for <math>P(n)</math>. The chance that she stops on the second box (<math>n=2</math>) is the chance of drawing a white marble then a red marble: <math>\frac{1}2 * \frac{1}3</math>. The chance that she stops on the third box (<math>n=3</math>) is the chance of drawing two white marbles then a red marble:<math>\frac{1}2 * \frac{2}3 * \frac{1}4</math>. If <math>n=4</math>, <math>P(n) = \frac{1}2 * \frac{2}3 * \frac{3}4 * \frac{1}5</math>.
  
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Cross-cancelling in the fractions gives <math>P(2) = </math>\frac{1}{2*3}<math>, </math>P(3) = <math>\frac{1}{3*4}</math>, and <math>P(4) = \frac{1}{4*5}</math>. From this, it is clear that <math>P(n) = \frac{1}{(n)(n+1)}</math>. (Alternatively, <math>P(n) = \frac{(n-1)!}{(n+1)!}</math>.)
+
Cross-cancelling in the fractions gives <math>P(2) = \frac{1}{2*3}</math>, <math>P(3) = </math>\frac{1}{3*4}<math>, and </math>P(4) = \frac{1}{4*5}<math>. From this, it is clear that </math>P(n) = \frac{1}{(n)(n+1)}<math>. (Alternatively, </math>P(n) = \frac{(n-1)!}{(n+1)!}<math>.)
  
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<math>\frac{1}{(n+1)(n)} < \frac{1}{2010}</math>
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</math>\frac{1}{(n+1)(n)} < \frac{1}{2010}<math>
  
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The lowest integer that satisfies the above inequality is <math>\boxed{(A) 45}</math>.
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The lowest integer that satisfies the above inequality is </math>\boxed{(A) 45}$.
  
 
== See also ==
 
== See also ==

Revision as of 18:09, 18 February 2013

Problem

Each of $2010$ boxes in a line contains a single red marble, and for $1 \le k \le 2010$, the box in the $k\text{th}$ position also contains $k$ white marbles. Isabella begins at the first box and successively draws a single marble at random from each box, in order. She stops when she first draws a red marble. Let $P(n)$ be the probability that Isabella stops after drawing exactly $n$ marbles. What is the smallest value of $n$ for which $P(n) < \frac{1}{2010}$?

$\textbf{(A)}\ 45 \qquad \textbf{(B)}\ 63 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ 201 \qquad \textbf{(E)}\ 1005$

Solution

Solution 1

The probability of drawing a white marble from box $k$ is $\frac{k}{k+1}$. The probability of drawing a red marble from box $n$ is $\frac{1}{n+1}$.

The probability of drawing a red marble at box $n$ is therefore

$\frac{1}{n+1} \left( \prod_{k=1}^{n-1}\frac{k}{k+1} \right) < \frac{1}{2010}$

$\frac{1}{n+1} \left( \frac{1}{n} \right) < \frac{1}{2010}$

$(n+1)n > 2010$

It is then easy to see that the lowest integer value of $n$ that satisfies the inequality is $\boxed{45\ \textbf{(A)}}$.

Solution 2

Using the first few values of $n$, it is easy to derive a formula for $P(n)$. The chance that she stops on the second box ($n=2$) is the chance of drawing a white marble then a red marble: $\frac{1}2 * \frac{1}3$. The chance that she stops on the third box ($n=3$) is the chance of drawing two white marbles then a red marble:$\frac{1}2 * \frac{2}3 * \frac{1}4$. If $n=4$, $P(n) = \frac{1}2 * \frac{2}3 * \frac{3}4 * \frac{1}5$.

Cross-cancelling in the fractions gives $P(2) = \frac{1}{2*3}$, $P(3) =$\frac{1}{3*4}$, and$P(4) = \frac{1}{4*5}$. From this, it is clear that$P(n) = \frac{1}{(n)(n+1)}$. (Alternatively,$P(n) = \frac{(n-1)!}{(n+1)!}$.)$\frac{1}{(n+1)(n)} < \frac{1}{2010}$The lowest integer that satisfies the above inequality is$\boxed{(A) 45}$.

See also

2010 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions
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